Question

evaluate integration of √(x)+ 1/ x + √(x) dx​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Answer:-

We have to evaluate:

$\displaystyle \sf \int{\dfrac{ \sqrt{x} + 1 }{x + \sqrt{x} } } \: dx$

Taking √x common in denominator we get;

$\implies \displaystyle \int \sf \frac{ \cancel{(\sqrt{x} + 1)}}{ \sqrt{x} \cancel{( \sqrt{x} + 1)} } \: dx \\ \\ \\ (\because \sf \sqrt{a} \times \sqrt{a} = a ) \\ \\ \\ \implies \displaystyle \int \sf \: \frac{1}{ {x}^{ \frac{1}{2} } } \: dx\\ \\ \\ ( \because \sf \sqrt[n]{x} = {x}^{ \frac{1}{n} } )$

Using 1/aⁿ = a⁻ⁿ we get;

$\: \implies \displaystyle \int \sf \: {x}^{ \frac{ - 1}{2} } dx$

Now;

using ∫ xⁿ dx = (xⁿ⁺¹) / n + 1 we get;

$\implies \sf \: \frac{ {x}^{ \frac{ - 1}{2} + 1} }{ \frac{ - 1}{2} + 1} + c \\ \\ \\ \implies \sf \: {x}^{ \frac{1}{2} } \times \frac{1}{ \frac{1}{2} } + c \\ \\ \\ \implies \sf \: \sqrt{x} \times 2 + c \\ \\ \\ \implies \underline{ \underline{ \sf \:2 \sqrt{x} + c}}$